Timeline for Alternative proofs sought after for a certain identity
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Dec 20, 2020 at 9:34 | comment | added | Martin Rubey | I find your answer very good, I certainly did not downvote it. English is not my first language, all I intended to say is that the two answers seem to be related. Put differently, also Fedor's proof seems to be your's in disguise. But in fact, I didn't try to make the relationship precise. (I also sometimes get downvotes I don't understand. It might happen by accident.) | |
| Dec 20, 2020 at 0:18 | comment | added | Iosif Pinelis | @MartinRubey : Thank you for uncovering the connection with Fedor Petrov's answer. However, I certainly did not try to disguise anything. (Is that why my answer got the down vote?) In fact, my answer was given before Fedor's. | |
| Dec 18, 2020 at 22:37 | comment | added | Martin Rubey | This can be rephrased combinatorially in a nice way: $\binom{n}{k}(k-1)!$ is the number of cyclic orderings of a $k$-element subset of $[n]$, so the first polynomial is the exponential generating function for cyclic orderings of subsets of $[n]$. On the other hand, $\frac{(1+t)^m-1}{m}$ is the exponential generating function for cyclic orders of subsets of $[m]$ whose largest element is $m$. I think that this is Fedor's proof in disguise. | |
| Dec 18, 2020 at 17:53 | comment | added | Iosif Pinelis | @TerryTao : Thank you for your comment -- this is a good way to put it. | |
| Dec 18, 2020 at 17:20 | comment | added | Terry Tao | To put it another way: $\sum_{k=1}^n \binom{n}{k} \frac{1}{k} = \sum_{k=1}^n \frac{2^k-1}{k}$ is a special case of $\sum_{k=1}^n \binom{n}{k} \frac{t^k}{k} = \sum_{k=1}^n \frac{(1+t)^k-1}{k}$ which is the antiderivative of $\sum_{k=1}^n \binom{n}{k} t^{k-1} = \sum_{k=1}^n (1+t)^{k-1}$, which follows from a combination of the binomial and geometric series identities. | |
| Dec 18, 2020 at 16:54 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |